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(D) For the function $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at the transition points $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{

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$-1, 1$

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(D) For the function $f(x)$ to be continuous in $[-\pi, \pi]$,it must be continuous at the transition points $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.At $x = -\frac{\pi}{2}$:$\lim_{x \rightarrow -\frac{\pi}{2}^-} (-2 \sin x) = -2 \sin(-\frac{\pi}{2}) = -2(-1) = 2$.$\lim_{x \rightarrow -\frac{\pi}{2}^+} (a \sin x + b) = a \sin(-\frac{\pi}{2}) + b = -a + b$.Equating these,we get $-a + b = 2$ (Equation $1$).At $x = \frac{\pi}{2}$:$\lim_{x \rightarrow \frac{\pi}{2}^-} (a \sin x + b) = a \sin(\frac{\pi}{2}) + b = a + b$.$\lim_{x \rightarrow \frac{\pi}{2}^+} \cos x = \cos(\frac{\pi}{2}) = 0$.Equating these,we get $a + b = 0$ (Equation $2$).Adding Equation $1$ and Equation $2$:$(-a + b) + (a + b) = 2 + 0 \Rightarrow 2b = 2 \Rightarrow b = 1$.Substituting $b = 1$ into Equation $2$:$a + 1 = 0 \Rightarrow a = -1$.Thus,the values are $a = -1$ and $b = 1$.

The values of $a$ and $b$ such that the function $f(x) = \begin{cases} -2 \sin x, & -\pi \leq x \leq -\frac{\pi}{2} \\ a \sin x + b, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$ is continuous in $[-\pi, \pi]$,are

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